E (HN/C) 4.5 6.5 X (nm) -4 Figure 1: 1. The electric field, E, as a function of a for a certain charge distribution is shown in figure 1. (a) The force that a positive point charge, q = 1.6 x 10-19C, experiences at the following positions, (i)= = 0 , (ii)= = 1am and (iii)= = 4nm (i) F1 = 0 (True, False) (ii) F2 = (2 N/C)(1.6 x 10-19() = 3.2 x 10 25 N (to the left) (True, False) (iii) Fa = (4AN/C)(1.6 x 10-190) = 6.4 x 10-25N (to the left) (True, False) (b) Determine the work done on this positive point charge between (i) a = 0 to a = 2nm (ii): = 2nm to a = 4.5am (iii)z = 4.5am to = = 6.5nm (iv)= = 0 to r = 6.5mm (1) W1 = q x area = (1.6 x 10-19c)( (-dAN/C) x (2nm) = -6.4 x 10-34 J (True, False) (ii) W2 = q x area = (1.6 x 10-190)((-4AN/C) x ((4.5 - 2)nm) = -16 x 10-34 J (True, False) (iii) Wa = q x area = (1.6 x 10-19c)( (-4#N/C) x ((6.5-4.5)nm) = _6.4 x 10-34 J (True, False) (iv) WA = -28.8 x 10-34J (True, False) (c) Determine the work done per unit positive point charge for the above intervals[potential differences]: (i) Vo - V2 = 1 = -4 x 10 15J/C= -4 x 10 15Volts (True, False) (ii) V2 - VA.s = 2 = -10 x 10 15J/C = -10 x 10 15 Volts (True,False) (iii) VA.5 - V6.5 = q WA = -4 x 10 15 J/C = -4 x 10-15Volts (True, False) (iv) Vo - Ve.s = WA = -4 x 10-15J/C = -18 x 10 15 Volts (True, False)